• Solve for $[x]$. (This helps because $[x]$ is always integer.)
• Convert the equation into two inequalities. Separate the cases for $[x]$ positive and negative.

Solution for $x$ is $x\in (-45,-\frac{2024}{45}]$.

Proof:

Let $[x] = \lfloor x\rfloor$ and $\{x\} = x - [x] \implies [x] = x-\{x\}$

$\implies 2025 > [x]^2+[x]\{x\} \geq 2024$

Case 1: 

$[x]$ is positive.

$[x]$ is an integer and, $0\leq \{x\}<1 \implies 0\leq \{x\}[x]<[x]$

$[[x[x]] = 2024 \implies [[x]^2+\{x\}[x]] = 2024$

From the first part,

$2025 >[x]^2+[x]\{x\}\implies 2025 > [x]^2 \implies [x]<45$

From the second part,

$[x]^2+[x]\{x\} \geq 2024\implies [x]^2 + [x] \geq 2024 \implies [x]>44$

So, $45>[x]>44$, no integer solution for $[x]$

Case 2:

$[x]$ is negative.

$[x]$ is an integer and $0\leq \{x\}<1 \implies 0\geq \{x\}[x]>[x]$

$[x[x]] = 2024 \implies [[x]^2+\{x\}[x]] = 2024$

From the first part,

$2025 >[x]^2+[x]\{x\}\implies 2025 > [x]^2+[x] \implies [x]>-46$

From the second part,

$[x]^2+[x]\{x\} \geq 2024\implies [x]^2 \geq 2024 \implies [x]<-44$

So, $-44>[x]>-46$ and so $[x] = -45$.

So, \[[-45x]=2024 \implies 2025 > -45x \geq 2024 \implies -\frac{2024}{45} \geq x > -45\]

So, the only solutions for $x$ is: $x\in (-45,-\frac{2024}{45}]$